Precalculus Examples

$\frac{x + 6}{x^{2} - 6 x + 8} > \frac{2}{x - 4}$

Step 1

Subtract $\frac{2}{x - 4}$ from both sides of the inequality.

$\frac{x + 6}{x^{2} - 6 x + 8} - \frac{2}{x - 4} > 0$

Step 2

Simplify $\frac{x + 6}{x^{2} - 6 x + 8} - \frac{2}{x - 4}$ .

Tap for more steps...

Step 2.1

Factor $x^{2} - 6 x + 8$ using the AC method.

Tap for more steps...

Step 2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $8$ and whose sum is $- 6$ .

$- 4, - 2$

Step 2.1.2

Write the factored form using these integers.

$\frac{x + 6}{(x - 4) (x - 2)} - \frac{2}{x - 4} > 0$

Step 2.2

To write $- \frac{2}{x - 4}$ as a fraction with a common denominator, multiply by $\frac{x - 2}{x - 2}$ .

$\frac{x + 6}{(x - 4) (x - 2)} - \frac{2}{x - 4} \cdot \frac{x - 2}{x - 2} > 0$

Step 2.3

Multiply $\frac{2}{x - 4}$ by $\frac{x - 2}{x - 2}$ .

$\frac{x + 6}{(x - 4) (x - 2)} - \frac{2 (x - 2)}{(x - 4) (x - 2)} > 0$

Step 2.4

Combine the numerators over the common denominator.

$\frac{x + 6 - 2 (x - 2)}{(x - 4) (x - 2)} > 0$

Step 2.5

Simplify the numerator.

Tap for more steps...

Step 2.5.1

Apply the distributive property.

$\frac{x + 6 - 2 x - 2 \cdot - 2}{(x - 4) (x - 2)} > 0$

Step 2.5.2

Multiply $- 2$ by $- 2$ .

$\frac{x + 6 - 2 x + 4}{(x - 4) (x - 2)} > 0$

Step 2.5.3

Subtract $2 x$ from $x$ .

$\frac{- x + 6 + 4}{(x - 4) (x - 2)} > 0$

Step 2.5.4

Add $6$ and $4$ .

$\frac{- x + 10}{(x - 4) (x - 2)} > 0$

Step 2.6

Factor $- 1$ out of $- x$ .

$\frac{- (x) + 10}{(x - 4) (x - 2)} > 0$

Step 2.7

Rewrite $10$ as $- 1 (- 10)$ .

$\frac{- (x) - 1 (- 10)}{(x - 4) (x - 2)} > 0$

Step 2.8

Factor $- 1$ out of $- (x) - 1 (- 10)$ .

$\frac{- (x - 10)}{(x - 4) (x - 2)} > 0$

Step 2.9

Rewrite $- (x - 10)$ as $- 1 (x - 10)$ .

$\frac{- 1 (x - 10)}{(x - 4) (x - 2)} > 0$

Step 2.10

Move the negative in front of the fraction.

$- \frac{x - 10}{(x - 4) (x - 2)} > 0$

Step 3

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x - 10 = 0$

$x - 4 = 0$

$x - 2 = 0$

Step 4

Add $10$ to both sides of the equation.

$x = 10$

Step 5

Add $4$ to both sides of the equation.

$x = 4$

Step 6

Add $2$ to both sides of the equation.

$x = 2$

Step 7

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 10$

$x = 4$

$x = 2$

Step 8

Consolidate the solutions.

$x = 10, 4, 2$

Step 9

Find the domain of $- \frac{x - 10}{(x - 4) (x - 2)}$ .

Tap for more steps...

Step 9.1

Set the denominator in $\frac{x - 10}{(x - 4) (x - 2)}$ equal to $0$ to find where the expression is undefined.

$(x - 4) (x - 2) = 0$

Step 9.2

Solve for $x$ .

Tap for more steps...

Step 9.2.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 4 = 0$

$x - 2 = 0$

Step 9.2.2

Set $x - 4$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 9.2.2.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 9.2.2.2

Add $4$ to both sides of the equation.

$x = 4$

Step 9.2.3

Set $x - 2$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 9.2.3.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 9.2.3.2

Add $2$ to both sides of the equation.

$x = 2$

Step 9.2.4

The final solution is all the values that make $(x - 4) (x - 2) = 0$ true.

$x = 4, 2$

Step 9.3

The domain is all values of $x$ that make the expression defined.

$(- \infty, 2) \cup (2, 4) \cup (4, \infty)$

Step 10

Use each root to create test intervals.

$x < 2$

$2 < x < 4$

$4 < x < 10$

$x > 10$

Step 11

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Tap for more steps...

Step 11.1

Test a value on the interval $x < 2$ to see if it makes the inequality true.

Tap for more steps...

Step 11.1.1

Choose a value on the interval $x < 2$ and see if this value makes the original inequality true.

$x = 0$

Step 11.1.2

Replace $x$ with $0$ in the original inequality.

$\frac{(0) + 6}{{(0)}^{2} - 6 \cdot 0 + 8} > \frac{2}{(0) - 4}$

Step 11.1.3

The left side $0.75$ is greater than the right side $- 0.5$ , which means that the given statement is always true.

True

Step 11.2

Test a value on the interval $2 < x < 4$ to see if it makes the inequality true.

Tap for more steps...

Step 11.2.1

Choose a value on the interval $2 < x < 4$ and see if this value makes the original inequality true.

$x = 3$

Step 11.2.2

Replace $x$ with $3$ in the original inequality.

$\frac{(3) + 6}{{(3)}^{2} - 6 \cdot 3 + 8} > \frac{2}{(3) - 4}$

Step 11.2.3

The left side $- 9$ is not greater than the right side $- 2$ , which means that the given statement is false.

False

Step 11.3

Test a value on the interval $4 < x < 10$ to see if it makes the inequality true.

Tap for more steps...

Step 11.3.1

Choose a value on the interval $4 < x < 10$ and see if this value makes the original inequality true.

$x = 7$

Step 11.3.2

Replace $x$ with $7$ in the original inequality.

$\frac{(7) + 6}{{(7)}^{2} - 6 \cdot 7 + 8} > \frac{2}{(7) - 4}$

Step 11.3.3

The left side $0.8\overline{6}$ is greater than the right side $0.\overline{6}$ , which means that the given statement is always true.

True

Step 11.4

Test a value on the interval $x > 10$ to see if it makes the inequality true.

Tap for more steps...

Step 11.4.1

Choose a value on the interval $x > 10$ and see if this value makes the original inequality true.

$x = 12$

Step 11.4.2

Replace $x$ with $12$ in the original inequality.

$\frac{(12) + 6}{{(12)}^{2} - 6 \cdot 12 + 8} > \frac{2}{(12) - 4}$

Step 11.4.3

The left side $0.225$ is not greater than the right side $0.25$ , which means that the given statement is false.

False

Step 11.5

Compare the intervals to determine which ones satisfy the original inequality.

$x < 2$ True

$2 < x < 4$ False

$4 < x < 10$ True

$x > 10$ False

$x < 2$ True

$2 < x < 4$ False

$4 < x < 10$ True

$x > 10$ False

Step 12

The solution consists of all of the true intervals.

$x < 2$ or $4 < x < 10$

Step 13

Convert the inequality to interval notation.

$(- \infty, 2) \cup (4, 10)$

Step 14